Normal Distribution: Car Journey

M

Monkeyseat

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Question:

Shanim drives from her home in Sale to college in Manchester every weekday during terms. On the way she collects her friend David who waits for her at the end of his road in Chorlton. Shamim leaves home at 8.00 a.m., and the time it takes her to reach the end of David's road is normally distributed with mean 23 minutes and standard deviation 5 minutres.

a) Find the probability that she arrives at the end of David's road before 8.30 a.m.

b) If David arrives at the end of is road at 8.05 a.m. what is the probability that he will have to wait less than 15 minutes for Shamim to arrive?

c) What is the latest time, to the nearest minute, that David can arrive at the end of his road to have a probability of at least 0.99 of arriving before Shamim?
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Working:

a) z value = (X - mean) / standard deviation

So I did (30-23)/5 which gave 1.4.

I went and looked that up in the normal distribution table and I got 0.91924 which the book says is correct.

b) z value = (X - mean) / standard deviation

So I did (20-23)/5 which gave -0.6.

Again, from the tables it gave me 0.72575. Because we are trying to find the probability of less than the negative z value, I did 1-0.72575 which gave 0.27425. The book also says this is correct.

c) I have no idea really what to do here. I think I have to work back to the z or X value but I'm not sure.
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So baisically, I'm stuck on (c). I've not been doing this for long so any help is appreciated (layman's terms if possible :)).

Thanks.
 
You can use \(\displaystyle z=\frac{x-{\mu}}{\sigma}\)

For the first one, \(\displaystyle \frac{30-23}{5}=1.4\)

Looking up a z-score of 1.4 in the table we see that we get 0.9192.

About a 92% probability she will arrive before 8:30.

See?. Now treat the others the same.
 
Monkeyseat said:
Working:

c) I have no idea really what to do here. I think I have to work back to the z or X value but I'm not sure.

What is the 'z' value for 99% on the low side?

So now find 99% minimum drive time and hence earliest arrival time for the car - using "backward" tranformation.
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galactus said:
You can use \(\displaystyle z=\frac{x-{\mu}}{\sigma}\)

For the first one, \(\displaystyle \frac{30-23}{5}=1.4\)

Looking up a z-score of 1.4 in the table we see that we get 0.9192.

About a 92% probability she will arrive before 8:30.

See?. Now treat the others the same.
Click to expand...

I had already done the same method you showed me for (a) and (b) as shown in the working in my first post, but don't know how to apply that to question (c). Thanks.

Subhotosh Khan said:
Monkeyseat said:
Working:

c) I have no idea really what to do here. I think I have to work back to the z or X value but I'm not sure.

What is the 'z' value for 99% on the low side?

So now find 99% minimum drive time and hence earliest arrival time for the car - using "backward" tranformation.
Click to expand...
Click to expand...
Click to expand...

To find the "z value for 99% on the low side" do I have to go through the table and look for the first one with 0.99....? Apologies, I do not know what you mean by this.

I think when I have the z value, I work back to X but would just like you to clarify the point above.

Thank you.

EDIT:

I just looked up 99% in the percentage points of the normal distribution table and got 2.3263. So I did:

z value = (X - mean)/standard deviation
2.3263 = (X - 23)/5
(5*2.3263) + 23 = X
X = 34.6315

I think I must have missed a step or done something wrong. Would you mind just giving me another hint?

Thanks.
 
I looked up 99% in the percentage points of the normal distribution table and got 2.3263. Should that be negative? When I tried it below with -2.3263 I got a similar answer to the book's:

z value = (X - mean)/standard deviation
-2.3263 = (X - 23)/5
(5*-2.3263) + 23 = X
X = 11.3685

If so, why should it be -2.3263? I thought it might be negative because you said "low side" - I took this as the left side of the normal distribution curve. However, I do not know why this is, or what you mean for certain, could you please clarify? I don't want you to give me the answer, I have tried to do it myself but just need another little it of help.

Much appreciated.

Thanks.
 
That looks good. He wants to arrive by about 8:12 to be 99% sure he beats her there.
 
galactus said:
That looks good. He wants to arrive by about 8:12 to be 99% sure he beats her there.
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Thank you for the reply galactus. Why is the z value for 99% -2.3263 and not just 2.3263 like I orginally thought though? When I looked 99% up in the table it said positive 2.3263. I thought what Subhotosh Khan said about "the low side" might have something to do with this, but not sure.

I'm just unsure about that, if you could just explain this I would be very grateful. Thanks.
 
galactus said:
You have to look up .01 in the table.
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Thanks for replying.

I thought it might be this because the normal distribution curve is symmetrical so that would give the negative value, but why exactly are we looking for 1% and not 99%?

Is it because we are looking for less than or equal to 1% (the probability of her being there before him) which is the same as finding greater than or equal to 99% (the probability of him being there before her)? If that is the case, is it correct that the z value exceeds 1% but is exceeded by 99%?

Positive 2.3263 would be the z value for the probability of LESS than 0.99 which is not what we are trying to find? I'm not entirely sure whether the assumptions I made are correct, I remember doing something vaguely similar in the past and I think I did something like I said above. Am I correct? if you could check for me what I said in this post that would be great and I would be very grateful. Again, many thanks.
 
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